Question: Is ${39254}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {39254}= &&{3}\cdot10000+ \\&&{9}\cdot1000+ \\&&{2}\cdot100+ \\&&{5}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {39254}= &&{3}(9999+1)+ \\&&{9}(999+1)+ \\&&{2}(99+1)+ \\&&{5}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {39254}= &&\gray{3\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {3}+{9}+{2}+{5}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${39254}$ is divisible by $3$ if ${ 3}+{9}+{2}+{5}+{4}$ is divisible by $3$ Add the digits of ${39254}$ $ {3}+{9}+{2}+{5}+{4} = {23} $ If ${23}$ is divisible by $3$ , then ${39254}$ must also be divisible by $3$ ${23}$ is not divisible by $3$, therefore ${39254}$ must not be divisible by $3$.